Cylindrical Triple Integral Find the Volume?

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SUMMARY

The discussion focuses on calculating the volume of a solid defined by the intersection of the cylinders described by the equations y² + z² = 1 and x² + z² = 1. The volume is determined using cylindrical coordinates and involves a double integral with specific limits. The final result of the volume calculation is confirmed to be 16, achieved through a series of transformations and integration techniques, including the application of L'Hôpital's Rule for evaluating limits.

PREREQUISITES
  • Understanding of cylindrical coordinates in multivariable calculus
  • Familiarity with double integrals and their applications
  • Knowledge of trigonometric identities and integration techniques
  • Experience with L'Hôpital's Rule for limit evaluation
NEXT STEPS
  • Study the conversion of Cartesian coordinates to cylindrical coordinates
  • Learn about evaluating double integrals in polar coordinates
  • Explore advanced integration techniques, including L'Hôpital's Rule
  • Investigate the geometric interpretation of volumes of solids defined by cylindrical boundaries
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Students and professionals in mathematics, particularly those studying calculus, as well as educators looking for examples of volume calculations in multivariable contexts.

jk8985
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Let E be the solid inside cylinder y^2+z^2=1 and x^2+z^2=1, find the volume of e and the surface area of e
 
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Can you show us what you have tried so our helpers know where you are stuck?
 
For the second half of the question, I got this. Is it correct?? Also, could you write it out nicely :( I don't really understand what I did.

Can someone let me know if this is correct and if I showed all steps?

Since the the region of integration inside x^2 + y^2 = 1 (and symmetry), convert to polar coordinates:
2 * 4 * ∫(θ = 0 to π/2) ∫(r = 0 to 1) (r dr dθ)/(1 - r^2 cos^2(θ))^(1/2)
= 8 * ∫(θ = 0 to π/2) ∫(r = 0 to 1) 2r (1 - r^2 cos^2(θ))^(-1/2) dr dθ
= 8 * ∫(θ = 0 to π/2) (-1/cos^2(θ)) 2(1 - r^2 cos^2(θ))^(1/2) {for r = 0 to 1} dθ
= 16 ∫(θ = 0 to π/2) (1/cos^2(θ)) [1 - (1 - cos^2(θ))^(1/2)] dθ
= 16 ∫(θ = 0 to π/2) (1 - sin θ) dθ/cos^2(θ)
= 16 ∫(θ = 0 to π/2) (sec^2(θ) - sec θ tan θ) dθ
= 16 (tan θ - sec θ) {for θ = 0 to π/2}
= 16 (sin θ - 1)/cos θ {for θ = 0 to π/2}
= 16 (0 - (-1)), using L'Hopital's Rule as θ→ π/2-
= 16.
 

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